Question

if sin theta=3/5, evaluate cosec theta-cos theta / 2 cot theta if 3 cos theta=1/3. find the value of 6 sin^2 theta + tan^2 theta /4 cos theta
please answer quickly ​

Answer 1

☆ Question :

• If $\sf{sin\theta = \dfrac{3}{5}}$ , then evaluate :

$\sf{\dfrac{cosec\theta cos\theta}{2cot\theta}}$

• If $\sf{3cos\theta = \dfrac{1}{3}}$ , then then find the value of :

$\sf{\dfrac{6 sin^{2}\theta + tan^{2}\theta}{4cos\theta}}$

☆ To Find :

The value of the Given Trigonometrical Equations.

• $\sf{\dfrac{cosec\theta cos\theta}{2cot\theta}}$

• $\sf{\dfrac{6 sin^{2}\theta + tan^{2}\theta}{4cos\theta}}$

☆ Given :

• $\sf{sin\theta = \dfrac{3}{5}}$

• $\sf{3cos\theta = \dfrac{1}{3}}$

☆ We Know :

☞ Pythagoras theorem :

$\purple{\sf{\underline{\boxed{h^{2} = b^{2} + p^{2}}}}}$

Where ,

• h = Hypotenuse

• p = Height

• b = base

☆ Solution (i) :

☞ Concept :

According to the question , we have to find the value of the Equation , by the given value.

We Know that , $\sf{sin\theta = \dfrac{p}{h}}$ and here it is given that , $\sf{sin\theta = \dfrac{3}{5}}$ , so it can be written as ,

$\green{\sf{sin\theta = \dfrac{p}{h} = \dfrac{3}{5}}}$

From the above Equation , we can conclude that the height is 3 units and the Hypotenuse is 5 units.

Now , by Pythagoras theorem we can find the base .

By using the Pythagoras theorem Substituting the values in it , we get :

$\purple{\sf{h^{2} = p^{2} + b^{2}}} \\ \\ \\ \implies \sf{5^{2} = 3^{2} + b^{2}} \\ \\ \\ \implies \sf{5^{2} - 3^{2} = b^{2}}$

By square Rooting on both the sides , we get :

$\implies \sf{\sqrt{5^{2} - 3^{2}} = \sqrt{b^{2}}} \\ \\ \\ \implies \sf{\sqrt{5^{2} - 3^{2}} = b} \\ \\ \\ \implies \sf{\sqrt{25 - 9} = b} \\ \\ \\ \implies \sf{\sqrt{16} = b} \\ \\ \\ \implies \sf{4 = b} \\ \\ \\ \therefore \purple{\sf{b = 4}}$

Now by this, the Triplet formed is 3 , 4 , 5 .

☞ Analysis :

Since the base is 4 units , height is 3 units and hypotenuse is 5 units.

We Know that ,

• $\sf{cosec\theta = \dfrac{h}{p}}$

Hence , the value of $\sf{cosec\theta}$ is :

$\sf{cosec\theta = \dfrac{h}{p} = \dfrac{5}{3}}$

Thus , $\sf{cosec\theta = \dfrac{5}{3}}$ .

• $\sf{cos\theta = \dfrac{b}{h}}$

Hence , the value of $\sf{cos\theta}$ is :

$\sf{cosec\theta = \dfrac{b}{h} = \dfrac{4}{5}}$

Thus , $\sf{cos\theta = \dfrac{4}{5}}$ .

• $\sf{cot\theta = \dfrac{b}{p}}$

Hence , the value of $\sf{cot\theta}$ is :

$\sf{cot\theta = \dfrac{b}{p} = \dfrac{4}{3}}$

Thus , $\sf{cot\theta = \dfrac{4}{3}}$ .

Now by putting the value of $\sf{cot\theta}$ , $\sf{cosec\theta}$ and $\sf{cos\theta}$ in the Equation , we can find the required value.

☞ Calculation :

$\purple{\sf{\dfrac{cosec\theta cos\theta}{2cot\theta}}} \\ \\ \implies \sf{\dfrac{\dfrac{5}{3} \times \dfrac{4}{5}}{2 \times \dfrac{4}{3}}} \\ \\ \implies \sf{\dfrac{\dfrac{\not{5}}{3} \times \dfrac{4}{\not{5}}}{2 \times \dfrac{4}{3}}} \\ \\ \implies \sf{\dfrac{\dfrac{1}{3} \times 4}{\dfrac{8}{3}}} \\ \\ \implies \sf{\dfrac{\dfrac{4}{3}}{\dfrac{8}{3}}}$

$\implies \sf{\dfrac{4}{3} \times \dfrac{3}{8}} \\ \\ \implies \sf{\dfrac{\not{4}}{\not{3}}} \times \dfrac{\not{3}}{8} \\ \\ \implies \sf{\dfrac{1}{2}} \\ \\ \therefore \purple{\sf{\dfrac{1}{2}}}$

Hence , the value of

$\sf{\dfrac{cosec\theta cos\theta}{2cot\theta}}$ is $\sf{\dfrac{1}{2}}$

☆ Solution (ii) :

By using the Pythagoras theorem Substituting the values in it , we get :

$\purple{\sf{h^{2} = p^{2} + b^{2}}} \\ \\ \\ \implies \sf{9^{2} = h^{2} + 1^{2}} \\ \\ \\ \implies \sf{9^{2} - 1^{2} = h^{2}}$

By square Rooting on both the sides , we get :

$\implies \sf{\sqrt{9^{2} - 1^{2}} = \sqrt{h^{2}}} \\ \\ \\ \implies \sf{\sqrt{9^{2} - 1^{2}} = h} \\ \\ \\ \implies \sf{\sqrt{81 - 1} = h} \\ \\ \\ \implies \sf{\sqrt{80} = b} \\ \\ \\ \implies \sf{4\sqrt{5} = h} \\ \\ \\ \\ \therefore \purple{\sf{h = 4\sqrt{5}}}$

☞ Analysis :

Since the base is 1 units , height is 4√5 units and hypotenuse is 5 units.

• $\sf{sin\theta = \dfrac{p}{h} = \dfrac{4\sqrt{5}}{9}}$

• $\sf{tan\theta = \dfrac{p}{b} = 4\sqrt{5}}$

Calculation :

$\purple{\sf{\dfrac{6 sin^{2}\theta + tan^{2}\theta}{4cos\theta}}} \\ \\ \\ \\ \implies \sf{\dfrac{6 \times \left(\dfrac{4\sqrt{5}}{9}\right)^{2} + \left(4\sqrt{5}\right)^{2}}{4 \times \dfrac{1}{9}}} \\ \\ \\ \\ \implies \sf{\dfrac{6 \times \dfrac{80}{81} + 80}{\dfrac{4}{9}}} \\ \\ \\ \\ \implies \sf{\dfrac{2 \times \dfrac{80}{27} + 80}{\dfrac{4}{9}}} \\ \\ \\ \\ \implies \sf{\dfrac{\dfrac{160}{27} + 80}{\dfrac{4}{9}}} \\ \\ \\ \\ \implies \sf{\dfrac{\dfrac{160 + 2160}{27}}{\dfrac{4}{9}}}$

$\implies \sf{\dfrac{\dfrac{2320}{27}}{\dfrac{4}{9}}} \\ \\ \\ \\ \implies \sf{\dfrac{2320}{27} \times \dfrac{9}{4}} \\ \\ \\ \\ \implies \sf{\dfrac{580}{3}} \\ \\ \\ \\ \implies \sf{193.33} \\ \\ \\ \\ \therefore \purple{\sf{193.33}}$

Hence , the value of $\sf{\dfrac{6 sin^{2}\theta + tan^{2}\theta}{4cos\theta}}$ is $\sf{193.33}$

Answer 2

Step-by-step explanation:

Solution (i) :

☞ Concept :

According to the question , we have to find the value of the Equation , by the given value.

We Know that , \sf{sin\theta = \dfrac{p}{h}}sinθ=

h

p

and here it is given that , \sf{sin\theta = \dfrac{3}{5}}sinθ=

5

3

, so it can be written as ,

\green{\sf{sin\theta = \dfrac{p}{h} = \dfrac{3}{5}}}sinθ=

h

p

=

5

3

From the above Equation , we can conclude that the height is 3 units and the Hypotenuse is 5 units.

Now , by Pythagoras theorem we can find the base .

By using the Pythagoras theorem Substituting the values in it , we get :

\begin{gathered}\purple{\sf{h^{2} = p^{2} + b^{2}}} \\ \\ \\ \implies \sf{5^{2} = 3^{2} + b^{2}} \\ \\ \\ \implies \sf{5^{2} - 3^{2} = b^{2}} \\ \\ \\\end{gathered}

h

2

=p

2

+b

2

⟹5

2

=3

2

+b

2

⟹5

2

−3

2

=b

2

By square Rooting on both the sides , we get :

\begin{gathered}\implies \sf{\sqrt{5^{2} - 3^{2}} = \sqrt{b^{2}}} \\ \\ \\ \implies \sf{\sqrt{5^{2} - 3^{2}} = b} \\ \\ \\ \implies \sf{\sqrt{25 - 9} = b} \\ \\ \\ \implies \sf{\sqrt{16} = b} \\ \\ \\ \implies \sf{4 = b} \\ \\ \\ \therefore \purple{\sf{b = 4}}\end{gathered}

⟹

5

2

−3

2

=

b

2

⟹

5

2

−3

2

=b

⟹

25−9

=b

⟹

16

=b

⟹4=b

∴b=4

Now by this, the Triplet formed is 3 , 4 , 5 .

☞ Analysis :

Since the base is 4 units , height is 3 units and hypotenuse is 5 units.

We Know that ,

\sf{cosec\theta = \dfrac{h}{p}}cosecθ=

p

h

Hence , the value of \sf{cosec\theta}cosecθ is :

\sf{cosec\theta = \dfrac{h}{p} = \dfrac{5}{3}}cosecθ=

p

h

=

3

5

Thus , \sf{cosec\theta = \dfrac{5}{3}}cosecθ=

3

5

.

\sf{cos\theta = \dfrac{b}{h}}cosθ=

h

b

Hence , the value of \sf{cos\theta}cosθ is :

\sf{cosec\theta = \dfrac{b}{h} = \dfrac{4}{5}}cosecθ=

h

b

=

5

4

Thus , \sf{cos\theta = \dfrac{4}{5}}cosθ=

5

4

.

\sf{cot\theta = \dfrac{b}{p}}cotθ=

p

b

Hence , the value of \sf{cot\theta}cotθ is :

\sf{cot\theta = \dfrac{b}{p} = \dfrac{4}{3}}cotθ=

p

b

=

3

4

Thus , \sf{cot\theta = \dfrac{4}{3}}cotθ=

3

4

.

Now by putting the value of \sf{cot\theta}cotθ , \sf{cosec\theta}cosecθ and \sf{cos\theta}cosθ in the Equation , we can find the required value.

☞ Calculation :

\begin{gathered}\purple{\sf{\dfrac{cosec\theta cos\theta}{2cot\theta}}} \\ \\ \implies \sf{\dfrac{\dfrac{5}{3} \times \dfrac{4}{5}}{2 \times \dfrac{4}{3}}} \\ \\ \implies \sf{\dfrac{\dfrac{\not{5}}{3} \times \dfrac{4}{\not{5}}}{2 \times \dfrac{4}{3}}} \\ \\ \implies \sf{\dfrac{\dfrac{1}{3} \times 4}{\dfrac{8}{3}}} \\ \\ \implies \sf{\dfrac{\dfrac{4}{3}}{\dfrac{8}{3}}} \\\end{gathered}

2cotθ

cosecθcosθ

⟹

2×

3

4

3

5

×

5

4

⟹

2×

3

4

3



5

×



5

4

⟹

3

8

3

1

×4

⟹

3

8

3

4

\begin{gathered}\implies \sf{\dfrac{4}{3} \times \dfrac{3}{8}} \\ \\ \implies \sf{\dfrac{\not{4}}{\not{3}}} \times \dfrac{\not{3}}{8} \\ \\ \implies \sf{\dfrac{1}{2}} \\ \\ \therefore \purple{\sf{\dfrac{1}{2}}}\end{gathered}

⟹

3

4

×

8

3

⟹



3



4

×

8



3

⟹

2

1

∴

2

1

Hence , the value of

\sf{\dfrac{cosec\theta cos\theta}{2cot\theta}}

2cotθ

cosecθcosθ

is \sf{\dfrac{1}{2}}

2

1